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EXPLANATION OF RHENIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 30 , 2015 ' ' INTRODUCTION Rhenium is a chemical element with symbol Re and atomic number 75. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Rhenium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25p6 4f145d1 5d15d1 5d15d16s2 According to the “Ionization Energies (eV) of Atoms and Ions” the ionization energies of rhenium ( from E1 to E5 ) are the following: E1 = 7.83 , E2 = 17 , E3 = 26 , E4 = 38 , and E5 = 51 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my new paper of 2008. EXPLANATION OF E1 = 7.83 eV = -E(6s2) + E(6s1) ' The electron charges (-73e) of the 73 electrons of the following configuration 1s22s22p63s23p63d104s24p64d105s25p6 4f 145d15d15d15d15d1 screen the nuclear charge (+75e) and for a perfect screening the electrons of 6s2 would provide an effective Zeff = ζ = 2. However the electrons of 6s2 penetrate the electrons of 5d5 with parallel spin and lead to the deformation of electron clouds. Therefore ζ > 2. Here the E(6s2) represents the binding energy of 6s2, while the E(6s1) represents the binding energy of 6s1, which appears after the first ionization of 6s2 . Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6s1 consists of one electron, we apply the Bohr formula as E(6s1) = (-13.6057)ζ2/n2 Therefore E1 = 7.83 eV = -E(6s2) + E(6s1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 287.86 = 0 and solving for ζ we get ζ = 5.18 > 2 . Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . ' ''' '''EXPLANATION OF E2 = 17 eV = -E(6s1) As in the case of E1 the electron charges (-73e) of the 73 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f145d15d15d15d15d1 screen the nuclear charge (+75e) and for a perfect screening we would have the same effective ζ = 2. However the one electron of 6s breaks the symmetry and leads to the greater deformation of electron clouds. Thus ζ > 5.18. Here the E(6s1) represents the binding energy of 6s1, given by applying the Bohr formula as E2 = 17 eV = E(6s1) = (-13.6057)ζ2/n2 Then using n = 6 we get ζ = 6.7 > 5.18. Here the ζ = 6.7 > 5.18 > 2 means that after the first ionization the one electron of 6s breaks the symmetry and leads to a greater deformation of electron clouds. ' ' EXPLANATION OF E3 = 26 eV = -E(5d1) The electron charges (-68e) of the 68 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f14) screen the nuclear charge (+75e) and for a perfect screening we would have an effective ζ = 7. Here the E(5d1) represents the binding energy of the first 5d1, given by applying the Bohr formula as E3 = 26 eV = E(5d1) = (-13.6057)ζ2/n2 Then using n = 5 we get ζ = 6.91 which is nearly equal to the perfect screening with ζ = 7. EXPLANATION OF E4 = 38 eV = -E(5d1) As in the case of E3 the electron charges (-68e) of the 68 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f 14) screen the nuclear charge (+75e) and for a perfect screening we would have the same effective ζ = 7. However after the first ionization of 5d the electrons of 5d break the symmetry and lead to the deformation of electron clouds. Thus ζ > 7 Here the E(5d1) represents the binding energy of the second 5d1, given by applying the Bohr formula as E4 = 38 eV = E(5d1) = (-13.6057)ζ2/n2 Then using n = 5 we get ζ = 8.36 > 7 Here the ζ = 8.36 > 7 means that after the ionization of the first electron of 5d the electrons of 5d break the symmetry and lead to a deformation of electron clouds. ' ' EXPLANATION OF E5 = 51 eV = -E(5d1) As in the case of E4 the electron charges (-68e) of the 68 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f14) screen the nuclear charge (+75e) and for a perfect screening we would have the same effective ζ = 7. However after the second ionization of 5d the electrons of 5d break more the symmetry and lead to the greater deformation of electron clouds. Thus ζ > 8.36 > 7. Here the E(5d1) represents the binding energy of the third 5d1, given by applying the Bohr formula as E5 = 51 eV = E(5d1) = (-13.6057)ζ2/n2 Then using n = 5 we get ζ = 9.68 > 8.36 > 7. Here the ζ = 9.68 > 8.36 > 7 means that after the ionizations the electrons of 5d break more the symmetry and lead to a greater deformation of electron clouds. Category:Fundamental physics concepts